JEE Main & Advanced Mathematics Quadratic Equations Question Bank Critical Thinking

  • question_answer 25) For what value of \[\lambda \]the sum of the squares of the roots of \[{{x}^{2}}+(2+\lambda )\,x-\frac{1}{2}(1+\lambda )=0\] is minimum   [AMU 1999]

    A) 3/2

    B) 1

    C) 1/2

    D) 11/4

    Correct Answer: C

    Solution :

    Given equation is \[{{x}^{2}}+(2+\lambda )x-\frac{1}{2}(1+\lambda )=0\] So, \[\alpha +\beta =-(2+\lambda )=0\] and \[\alpha \beta =-\left( \frac{1+\lambda }{2} \right)\] Now, \[{{\alpha }^{2}}+{{\beta }^{2}}={{(\alpha +\beta )}^{2}}-2\alpha \beta \] \[\Rightarrow {{\alpha }^{2}}+{{\beta }^{2}}={{\left[ -(2+\lambda ) \right]}^{2}}+2.\frac{(1+\lambda )}{2}\] Þ \[{{\alpha }^{2}}+{{\beta }^{2}}={{\lambda }^{2}}+4+4\lambda +1+\lambda ={{\lambda }^{2}}+5\lambda +5\] which is minimum for \[\lambda =1/2\].

adversite



LIMITED OFFER HURRY UP! OFFER AVAILABLE ON ALL MATERIAL TILL TODAY ONLY!

You need to login to perform this action.
You will be redirected in 3 sec spinner

Free
Videos